160. Intersection of Two Linked Lists（python）

Write a program to find the node at which the intersection of two singly linked lists begins.
For example, the following two linked lists:
A: a1 → a2
↘
c1 → c2 → c3
↗
B: b1 → b2 → b3
begin to intersect at node c1.
Notes:
If the two linked lists have no intersection at all, return null.
The linked lists must retain their original structure after the function returns.
You may assume there are no cycles anywhere in the entire linked structure.
Your code should preferably run in O(n) time and use only O(1) memory.
总结：关键点在于两链表的长度m、n(m>=n),长链表比短链表多走m-n步，这样两链表长度对应后，逐一比较就可以
Runtime: 392 ms
总结:无相同值时，返回空为None

# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
    def getLinklistlen(self,head):
        length=0
        while head:
            length=length+1
            head=head.next
        return length

    def getIntersectionNode(self, headA, headB):

        dummyA=ListNode(0)
        dummyA.next=headA
        dummyB=ListNode(0)
        dummyB.next=headB

        lenA=self.getLinklistlen(headA)
        lenB=self.getLinklistlen(headB)
        if lenA>lenB:
            differlen=lenA-lenB
            for i in range(differlen):
                dummyA=dummyA.next
        if lenA<lenB:
            differlen=lenB-lenA
            for i in range(differlen):
                dummyB=dummyB.next
        while  dummyA.next and dummyB.next:
            if dummyA.next==dummyB.next:
                return dummyA.next
            dummyA=dummyA.next
            dummyB=dummyB.next
        return None